Question: A candy machine holds $500$ pieces of candy, $20\%$ of which are blue. Customers get an SRS of $4$ pieces of candy per purchase from this machine. Let $B=$ the number of blue candies a customer gets in a purchase from a full machine. Which of the following would find $P(B=1)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(0.20)(0.80)^3$ (Choice B) B ${500 \choose 4}(0.20)(0.80)^3$ (Choice C) C ${500 \choose 4}(0.20)^3(0.80)$ (Choice D) D ${4 \choose 1}(0.20)^3(0.80)$ (Choice E) E ${4 \choose 1}(0.20)(0.80)^3$
Probability of $1$ blue candy We want the probability that there is $1$ success (blue candy) in $4$ trials (number of candies), so we're going to need $3$ failures (not blue candies) as well. The probability of each success is ${0.20}$ and the probability of each failure is $0.80}$. Since we're sampling less than $10\%$ of the population, we can assume independence and multiply probabilities to find the probability of getting $1$ success followed by $3$ failures: $\begin{aligned} P(\text{SFFF})&=\left({0.20}\right)\left(0.80}\right)\left(0.80}\right)\left(0.80}\right) \\\\ &=\left({0.20}\right)\left(0.80}\right)^3 \end{aligned}$ The binomial coefficient ${n \choose k}$ SFFF isn't the only arrangement that produces $1$ success in $4$ trials. For instance, FFFS would also produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=1$ success (blue candy) in $n=4$ trials (number of candies), so we should use the binomial coefficient ${4 \choose 1}$. [Tell me more about the binomial coefficient.] Putting it together Each arrangement has probability $(0.20)(0.80)^3$ so for our final answer we multiply this probability by the number of possible arrangements: ${4 \choose 1}(0.20)(0.80)^3$ The answer: ${4 \choose 1}(0.20)(0.80)^3$